175- Partial fractions decomposition

November 18, 2008

[This post replaces the previous version from October 12, 2008. The new argument is significantly simpler than the one originally posted.]

I want to present here a Calculus II-level proof that the method of partial fractions decomposition works. I will actually show a more general result, which will greatly simplify the presentation and get rid of most problems of the somewhat awkward formulation below.

We need some notation:

  • \prod_{i=1}^n a_i means a_1\times a_2\times\dots\times a_n.
  • p(x), q(x), etc, denote polynomials with real coefficients.

The proof that I show below is algorithmic in nature, meaning that it provides us with a method to find the relevant constants for any given specific polynomials p and q. The constants we obtain are real numbers.

That the method of partial fractions decomposition works means that we can always find the relevant constants the method requires.

Read the rest of this entry »


175 -A problem from Homework sets 9, 10

November 18, 2008

I want to show here how to solve the following problem from this week’s homework set:

Starting with a given x_0, define the subsequent terms of a sequence by setting x_{n+1}=x_n+\sin(x_n). Determine whether the sequence \{x_n\} converges, and if it does, find its limit.

This is a nice simple example of a (discrete) dynamical system in one variable. It turns out that the sequence always converges, but the limit depends on the value of x_0.

  • Case 1. x_0=n\pi for some n=0,\pm1,\pm2,\dots

In this case x_1=x_2=\dots=n\pi, so the sequence trivially converges.

  • Case 2. 2n\pi<x_0<(2n+1)\pi for some n=0,\pm1,\pm2,\dots

In this case I will show that x_0<x_1<\dots and that \lim_{i\to\infty}x_i=(2n+1)\pi.

First, notice that if 2n\pi<x_i<(2n+1)\pi, then we can write x_i=2n\pi+t for some t with 0<t<\pi and \sin(x_i)=\sin(t)>0, so x_{i+1}=x_i+\sin(x_i)>x_i.

Second, recall that \sin\theta<\theta for all \theta>0. You are probably familiar with this inequality from Calculus I; if not, one can prove it easily as follows: Let f(\theta)=\sin(\theta)-\theta, so f(0)=0. Also, f'(\theta)=\cos(\theta)-1\le0 for all \theta, so f is always decreasing, and the result follows.

Also, recall that \sin(t)=\sin(\pi-t), so

x_{i+1}=x_i+\sin(x_i)=2n\pi+t+\sin(t) =2n\pi+t+\sin(\pi-t)<2n\pi+t+(\pi-t)=(2n+1)\pi.

We have shown (by induction) that the sequence \{x_i\}_{i\ge0} is strictly increasing and bounded above (by (2n+1)\pi). Thus, it converges. If L is its limit, then


so \sin(L)=0 and since 2n\pi<L\le(2n+1)\pi, it follows that L=(2n+1)\pi.

  • Case 3. (2n+1)\pi<x_0<(2n+2)\pi for some n=0,\pm1,\pm2,\dots

In this case, x_0>x_1>\dots and \lim_{i\to\infty}x_i=(2n+1)\pi.

The argument is very similar to the one for Case 2: If (2n+1)\pi<x_i<(2n+2)\pi, then \sin(x_i)<0 so x_{i+1}<x_i, and x_i=(2n+1)\pi+t for some t\in(0,\pi), so \sin(x_i)=-\sin(t)>-t and therefore x_{i+1}>(2n+1)\pi. It follows that the sequence \{x_i\}_{i\ge0} is decreasing and bounded below (by (2n+1)\pi), so it converges. As before, the limit must in fact be (2n+1)\pi, and we are done.

175, 275 -Homework 11 and suggestions for the week after Thanksgiving

November 18, 2008

Homework 11 is due Tuesday, December 2, at the beginning of lecture. The usual considerations apply.

In 175 we will try to cover this week until section 8.10, but probably won’t get that far. As mentioned last week,  we will also cover additional topics that the book doesn’t mention or doesn’t treat in sufficient detail, once we are done with 8.10. These topics are uniform vs. pointwise convergence (including Wierstrass test), the behavior of the p-series \displaystyle \sum_{n=1}^\infty\frac1{n^p} for p=2,3,\dots, and infinite products. I will distribute notes of the topics not covered in the book. If you want to read ahead, and would like some of the notes ahead of time, please let me know.

In 275 we will cover Chapter 14, probably this week we will cover until section 14.4. Particularly important are the notion of conservative field and Green’s theorem. Then we will continue with surface integrals and orientations, Stokes’s and the divergence theorems. This will probably take another week, maybe a bit more. If there is any time left afterwards, we will see Lagrange multipliers 

Homework 11:

175: Do not use the solutions manual for any of these problems.

  • Turn in the problems listed for Homework 10  that you still have pending.
  • Section 8.6. Exercises 8, 25, 28, 37, 60.
  • Section 8.7. Exercises 2, 4, 39-48. 

Besides the exercises you have pending from last week, there are 17 new problems. Turn in the exercises you have pending, and at least 10 of the new problems. The others (at most 7) will be due December 9, together with the additional exercises for that week. 


  • Turn in the problems listed for Homework 10  that you still have pending.
  • Section 14.1. Exercises 1-8, 12, 16, 29.
  • Section 14.2. Exercises 6, 8, 34, 41.
  • Section 14.3. Exercises 2, 6, 12, 17, 20, 34, 38.
  • Section 14.4. Exercises 2, 8, 18, 31-35.

Exercises 14.1.1-8 count as a single exercise. Besides the problems pending from last week, there are 23 exercises. Turn in at least 10 of these. The remaining problems (at most 13) will be due together with a few additional exercises on December 9.

A characterization of continuity

November 18, 2008

Yesterday, Randall Holmes mentioned to me the following nice characterization of continuity for functions between Euclidean spaces.

Theorem. A function f:{mathbb R}^nto{mathbb R}^m is continuous iff it preserves path-connectedness and compactness.

This is an easy exercise, but I don’t remember having seen the characterization before, so I figured I could as well write down the argument I found. It is clear that the result holds for a much wider class of spaces than the {mathbb R}^n, but to keep this post simple, I’ll just leave it this way.

Proof. For Euclidean spaces, continuity and sequential continuity coincide. Towards a contradiction, assume x_i converges to x but f(x_i) does not converge to f(x).

  • Case 1. The range of {f(x_i): ige 0} is infinite.

This quickly leads to a contradiction since {f(x_i):ige 0}cup{f(x)} is a compact set: We may as well assume that the map imapsto f(x_i) is injective, and since f(x_i) does not converge to f(x) we may in fact assume that all the f(x_i) stay away from f(x). The set {f(x_i):ige 0} has an accumulation point, which cannot be f(x) so it must be f(x_m) for some m. But then the set {f(x_i):i>m}cup{f(x)} is both compact and lacks one of its accumulation points, contradiction.

  • Case 2. The range is finite.

We may as well assume all x_i have the same image. Fix paths A_i=[x_i,x_{i+1}] in {mathbb R}^n that we can combine to get a path bigcup_i A_icup{x} (for example, A_i could simply be a segment). By preservation of path connectedness, any A_i with range of size at least 2 in fact has range of size continuum. If infinitely many of the A_i have infinite range, one easily reduces to Case 1. So we may assume all the A_i have constant range, but then bigcup_i A_icup{x} has a disconnected image. {sf QED}